Problem: Solve for $x$ : $ 8|x - 3| - 4 = -6|x - 3| + 8 $
Solution: Add $ {6|x - 3|} $ to both sides: $ \begin{eqnarray} 8|x - 3| - 4 &=& -6|x - 3| + 8 \\ \\ { + 6|x - 3|} && { + 6|x - 3|} \\ \\ 14|x - 3| - 4 &=& 8 \end{eqnarray} $ Add ${4}$ to both sides: $ \begin{eqnarray} 14|x - 3| - 4 &=& 8 \\ \\ { + 4} &=& { + 4} \\ \\ 14|x - 3| &=& 12 \end{eqnarray} $ Divide both sides by ${14}$ $ \dfrac{14|x - 3|} {{14}} = \dfrac{12} {{14}} $ Simplify: $ |x - 3| = \dfrac{6}{7}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x - 3 = -\dfrac{6}{7} $ or $ x - 3 = \dfrac{6}{7} $ Solve for the solution where $x - 3$ is negative: $ x - 3 = -\dfrac{6}{7} $ Add ${3}$ to both sides: $ \begin{eqnarray} x - 3 &=& -\dfrac{6}{7} \\ \\ {+ 3} && {+ 3} \\ \\ x &=& -\dfrac{6}{7} + 3 \end{eqnarray} $ Change the ${ + 3}$ to an equivalent fraction with a denominator of $7$ $ x = - \dfrac{6}{7} {+ \dfrac{21}{7}} $ $ x = \dfrac{15}{7} $ Then calculate the solution where $x - 3$ is positive: $ x - 3 = \dfrac{6}{7} $ Add ${3}$ to both sides: $ \begin{eqnarray} x - 3 &=& \dfrac{6}{7} \\ \\ {+ 3} && {+ 3} \\ \\ x &=& \dfrac{6}{7} + 3 \end{eqnarray} $ Change the ${ + 3}$ to an equivalent fraction with a denominator of $7$ $ x = \dfrac{6}{7} {+ \dfrac{21}{7}} $ $ x = \dfrac{27}{7} $ Thus, the correct answer is $x = \dfrac{15}{7} $ or $x = \dfrac{27}{7} $.